

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Tree method}	%Opgave 1
		\begin{enumerate}[a)] 
			\item									%Opgaven 1a
				\textbf{Claim}. $A \rightarrow B, \neg C \rightarrow \neg B, \neg A \rightarrow A \models C $.
				
				\begin{proof}
	\space\
				\begin{figure}[!ht]
				\qtreepadding=3pt
						\Tree
						[.{%Hoofd tak
								\begin{tabular}{rll}
									1. &$ A \rightarrow B $							& Premiss 1								\\
									2. &$ \neg C \rightarrow \neg B $		& Premiss 2								\\
									3. &$ \neg A \rightarrow A $				& Premiss 3								\\
									4. &$ \neg C $											& $\neg$ Conclusion 1				\\
									5. &$ \neg A \vee B $								& $\rightarrow$ Rule 1			\\
									6. &$ \neg \neg C \vee \neg B $			&	$\rightarrow$ Rule 2			\\
									7. &$ \neg \neg A \vee A $					&	$\rightarrow$	Rule 3			\\
								\end{tabular}	
							}
							[.{
								\begin{tabularx}{100pt}{rXl}
									8. &$\neg \neg C$							& $\vee$ Rule 6
								\end{tabularx}
								}
								{$\otimes$ (4, 8)}
							]
							[.{
									\begin{tabularx}{100pt}{rXl}
										9. &$\neg B$								& $\vee$ Rule 6
									\end{tabularx}
								}
								[.{
										\begin{tabularx}{100pt}{rXl}
											10.&$\neg A$						& $\vee$ Rule 5
										\end{tabularx}
									}
									[.{
											\begin{tabularx}{100pt}{rXl}
												12. &$\neg \neg A$		& $\vee$ Rule 7
											\end{tabularx}
										}
										{$\otimes$ (10, 12)}
									]	
									[.{
											\begin{tabularx}{100pt}{rXl}
												13. &$A$						& $\vee$ Rule 7
											\end{tabularx}
										}
										{$\otimes$ (10, 13)}
									]
								]
								[.{
										\begin{tabularx}{100pt}{rXl}
											11.&$B$						& $\vee$ Rule 5
										\end{tabularx}
									}
									{
										$\otimes$ (9, 11)
									}
								]
							]
						]
						\caption{Tree of $A \rightarrow B, \neg C \rightarrow \neg B, \neg A \rightarrow A \models C $.}
						\label{figure:Boom1a}
				\end{figure}\space\\		
			The claim is correct because all branches of the tree closes and lead to a contradiction. Because the collection is not fullfillable, and the collection is the negation of the original claim, it is proven that the original claim must be correct.
\end{proof}
			
			
			\item 								%Opgaven 1b
				\textbf{Claim}: $\models ((A \rightarrow B) \rightarrow A) \rightarrow A $.
				\begin{proof}
				\space\
				\qtreepadding=3pt
				\begin{figure}[!ht]
						\Tree
						[.{%Hoofd tak
								\begin{tabular}{rll}
									1. &$ \neg ((A \rightarrow B) \rightarrow A) \rightarrow A) $		& $\neg$ Conclusion				\\
									2. &$ ((A \rightarrow B) \rightarrow A) \wedge \neg A $					& $\rightarrow$ Rule 1		\\
									3. &$ ((A \rightarrow B) \rightarrow A) $												&	$\wedge$ Rule 2				\\
									4. &$ \neg A$																										& $\wedge$ Rule 2				\\
									5. &$ \neg (A \rightarrow B ) \vee A$													& $\rightarrow$ Rule 3		\\
								\end{tabular}
							}
							[.{ %Sub boom
									\begin{tabularx}{150pt}{rXl}
										6. &$ \neg (A \rightarrow B) $														& $\vee$ Rule 5.						\\
										8. &$ A \wedge \neg B $																		& $\rightarrow$ Rule 6.		\\
										9. &$ \neg B $																						& $\wedge$ Rule 6.					\\
										10.&$ A$																									& $\wedge$ Rule 6.					\\
									\end{tabularx}	
								}
								{$\otimes$ (4, 10)}
							]
							[.{
								\begin{tabularx}{100pt}{rXl}
									7. & $A $						& $\vee$ Rule 5.				
								\end{tabularx}
								}
								{$\otimes$ (4, 7)}
							]
						]
						\caption{Tree of $\models ((A \rightarrow B) \rightarrow A) \rightarrow A $.}
						\end{figure}\space\\		
			The claim is correct because all branches of the tree closes and lead to a contradiction. Because the collection is not fullfillable, and the collection is the negation of the original claim, it is proven that the original claim must be correct.
\end{proof}
						
			\item 								%Opgaven 1c
				\begin{enumerate}[-i-] 
					\item 							%opgaven 1ci
						\textbf{Claim}: $ \neg (B \wedge \neg C), A \rightarrow B \models \neg C \rightarrow \neg A $
							\begin{proof}\space\
							\qtreepadding=3pt
							\begin{figure}[!ht]
								\Tree
								[.{%Hoofd tak
										\begin{tabular}{rll}
											1. &$ \neg (B \wedge \neg C) $ 					& Premiss 1 							\\
											2. &$ A \rightarrow B $ 								&	Premiss 2 							\\
											3. &$ \neg(\neg C \rightarrow \neg A) $ & $\neg$Conclusion 				\\
											4. &$ \neg B \vee \neg \neg C $ 				& $\neg$ Rule 1. 				\\
											5. &$ \neg A \vee B $										& $\rightarrow$ Rule 2.	\\
											6. &$ \neg C \wedge \neg \neg A $				&	$\rightarrow$ Rule 3.	\\
											7. &$ \neg C $													&	$\wedge$ Rule 6.				\\
											8. &$ \neg \neg A $											& $\wedge$ Rule 6.				\\
											9. &$ A $																& $\neg$ Relgel 8.
										\end{tabular}
									}
									[.{ %Sub boom
											\begin{tabularx}{100pt}{rXl}
													10. $\neg B$ &$\vee$ Rule 4.
											\end{tabularx}
										} 
										[.{ %Linker deel sub boom
											\begin{tabularx}{100pt}{rXl}
												12. $\neg A$ &$\vee$ Rule 5.
											\end{tabularx}
											}
											{$\otimes$ (9, 12)}
										]
										%Rechter deel sub boom
										[.{ 
												\begin{tabularx}{100pt}{rXl}
													13. $B$ 		&$\vee$ Rule 5.
												\end{tabularx}
											} 
											{$\otimes$ (10, 13)}
										]
									]	
									%rechter deel hoofd boom							
									[.{ 
										\begin{tabularx}{110pt}{rXl}
												11. $\neg \neg C$ &$\vee$ Rule 4.
										\end{tabularx}
									} 
									{$\otimes$ (7, 11)}
									]
								]
								\caption{Tree of  $ \neg (B \wedge \neg C), A \rightarrow B \models \neg C \rightarrow \neg A $.}
								\label{figure:Boom1ci}
						\end{figure}\space\\		
			The claim is correct because all branches of the tree closes and lead to a contradiction. Because the collection is not fullfillable, and the collection is the negation of the original claim, it is proven that the original claim must be correct.
\end{proof}
							
						\clearpage
					\item 							%opgaven 1cii
						\textbf{Claim}: $ A \vee \neg (B \wedge \neg C), A \rightarrow B \models \neg C \rightarrow \neg A$.
						\space\
						\qtreepadding=3pt
						\begin{figure}[!ht]
							\Tree
							[.{%Hoofd boom
									\begin{tabular}{rll}
										1. &$A \vee \neg(B \wedge \neg C)$							&Premiss 1						\\
										2. &$A \rightarrow B$														&Premiss 2						\\
										3. &$\neg ( \neg C \rightarrow \neg A)$					&$\neg$Conclusion 			\\ 
										4. &$\neg A \vee B$															&$\rightarrow$Rule 2	\\
										5. &$\neg C \wedge \neg \neg A$									&$\rightarrow$Rule 3	\\
										6. &$\neg C$																		&$\wedge$Rule 5			\\
										7. &$\neg \neg A$																&$\wedge$Rule 5			\\
										8. &$A$																					&$\neg$Rule 7				\\
									\end{tabular}
								}
								[.{
										\begin{tabularx}{80pt}{rXl}
											9. &$A$																				&$\vee$ Rule 1
										\end{tabularx}
									}
									[.{
											\begin{tabularx}{80pt}{rXl}
												14. &$\neg A$																&$\vee$ Rule 4
											\end{tabularx}
										}
										{$\otimes$ (8, 14)}
									]
									{
										\begin{tabularx}{80pt}{rXl}
											15. &$B$																			&$\vee$ Rule 4
										\end{tabularx}
									}
								]
								[.{
										\begin{tabularx}{140pt}{rXl}
											10. &$\neg ( B \wedge \neg C )$								&$\vee$ Rule 1				\\
											11. &$\neg B \vee \neg \neg C $								&$\wedge$ Rule 10			\\
										\end{tabularx}
									}
									[.{
											\begin{tabularx}{100pt}{rXl}
												12. &$\neg B $																&$\vee$ Rule 10
											\end{tabularx}
										}
										[.{
											\begin{tabularx}{90pt}{rXl}
												16. &$\neg A $																&$\vee$ Rule 4
											\end{tabularx}
											}
											{$\otimes$ (8, 16)}
										]
										[.{
												\begin{tabularx}{90pt}{rXl}
													17. &$B$																		&$\vee$ Rule 4
												\end{tabularx}
											}
											{$\otimes$ (12, 17)}
										]											
									]
									[.{
											\begin{tabularx}{100pt}{rXl}
												13. &$\neg \neg C $													&$\vee$ Rule 10
											\end{tabularx}
										}
										{$\otimes$ (6, 13)}
									]								]	
							]
							\caption{Tree of  $ A \vee \neg (B \wedge \neg C), A \rightarrow B \models \neg C \rightarrow \neg A$.}
							\label{figure:Boom1cii}
						\end{figure}\space\\		
						The claim is incorrect because not all the branches are closed (branche 15 does not) and a counter exemple can be direved from this branch.
						The valuations: $V(A)=V(B)=(V\neg C)$ - $A = 1, B = 1, C = 0$ form a counter example and thus the claim is not true.

					\item 							%opgave 1ciii
						The right side of the tree under points 10. and 11. in 1c.ii is the same as the tree from 1c.i. But as 1c.ii has an extra $A \vee$ in front of the first premiss, there is a extra brance that will not close and so the claim is not true in 1c.iii and true in 1c.ii
				\end{enumerate}
		\end{enumerate}
		
		
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%	
\clearpage
\section{Resolution method}		%Opgave 2

		\begin{enumerate}[a)] 
			\item									%Opgave 2a
			
				\textbf{Claim}: $ A \vee \neg (B \wedge \neg C), A \rightarrow B \models \neg C \rightarrow \neg A $.
			
				\begin{align*}
					A \vee  \neg( B \wedge \neg C ) & \equiv A \vee (\neg B \vee \neg \neg C)	\\
																					& \equiv A \vee \neg B \vee C		
				\end{align*}
				
				\begin{align*}
					A \rightarrow B & \equiv \neg A \vee B		\\
				\end{align*}
				
				\begin{align*}
					\neg ( \neg C \rightarrow \neg B) & \equiv \neg C \wedge \neg \neg A	\\
																						& \equiv \neg C \wedge A 
				\end{align*}
				
				$ F = \left\{A \vee \neg B \vee C, \neg A \vee B, \neg C, A\right\} $
				
				\begin{align*}
					&\text{1. } A \vee \neg B \vee C									&\text{Element of F}				\\
					&\text{2. } \neg A \vee B													&\text{Element of F}				\\
					&\text{3. } \neg C																&\text{Element of F}				\\
					&\text{4. } A																			&\text{Element of F}				\\
					&\text{5. } B																			&\text{Resolvent (2,4)}			\\
					&\text{6. } A \vee \neg B													&\text{Resolvent (1,3)}			\\
					&\text{6. } A																			&\text{Resolvent (5,6)}			\\	
				\end{align*}
			
			The claim is incorrect because a counterexample (the variables that remain in single)  has been deriven: $A=1, B=1, C=0$. 
			This is because if all premises are assumed true, the conclusion does not logically follow.
					
			\item 								%Opgave 2b
			
				\textbf{Claim}: $ (A \wedge B) \rightarrow \neg C, \neg (C \rightarrow D ), B \vee D \models \neg A $.
			\begin{proof}	
				\begin{align*}
					(A \wedge B ) \rightarrow \neg C & \equiv \neg ( A \wedge B ) \vee \neg C			\\
																					 & \equiv \neg A \vee \neg B \vee \neg C			\\
				\end{align*}
				
				\begin{align*}
					\neg ( C  \rightarrow D ) & \equiv C \wedge \neg D														\\
				\end{align*}
						
				$ F = \left\{ \neg A \vee \neg B \vee \neg C, C, \neg D, B \vee D, A\right\} $
				
				\begin{align*}
					&\text{1. } \neg A \vee \neg B \vee \neg C				&\text{Element of F}				\\
					&\text{2. } C																			&\text{Element of F}				\\
					&\text{3. } \neg D																&\text{Element of F}				\\
					&\text{4. } B \vee D															&\text{Element of F}				\\
					&\text{5. } A																			&\text{Element of F}				\\
					&\text{6. } B																			&\text{Resolvent (3,4)}			\\
					&\text{7. } \neg A \vee \neg B										&\text{Resolvent (1,2)}			\\			
					&\text{8. } \neg A																&\text{Resolvent (6,7)}			\\
					&\text{9. } \Box																	&\text{Resolvent (8,5)}			\\	
				\end{align*}
The claim is correct because $F$ is not fullfillable. In other words, a valuation for the variables in $F$ can be found to cause a contradiction. And because $F$ is the negation of the original claim, it is proven that the original claim must be correct.
\end{proof}				
			\item 								%Opgave 2c
				\textbf{Claim}: $ A \leftrightarrow B, (\neg A \wedge \neg B) \rightarrow \neg C, C \models B $.
			\begin{proof}	
				\begin{align*}
					A \leftrightarrow B & \equiv ( A \rightarrow B ) \wedge ( B \rightarrow A )			\\
															& \equiv ( \neg A \vee B ) \wedge ( \neg B \vee A)					\\
				\end{align*}
												
				\begin{align*}
					( \neg A \wedge \neg B ) \rightarrow \neg C & \equiv \neg ( \neg A \wedge \neg B ) \vee \neg C		\\
																										 & \equiv A \vee B \vee \neg C
				\end{align*}
				
				$ F = \left\{ \neg A \vee B, \neg B \vee A, A \vee B \vee \neg C, C, \neg B \right\} $
				
				\begin{align*}
					&\text{1. } \neg A \vee B														&\text{Element of F}				\\
					&\text{2. } \neg B \vee A 													&\text{Element of F}				\\
					&\text{3. } A \vee B \vee \neg C 										&\text{Element of F}				\\
					&\text{4. } C 																			&\text{Element of F}				\\
					&\text{5. } \neg B																	&\text{Element of F}				\\
					&\text{6. } A \vee B																&\text{Resolvent (3,4)}				\\
					&\text{7. } A 																			&\text{Resolvent (5,6)}				\\
					&\text{8. } B																				&\text{Resolvent (1,7)}				\\
					&\text{9. } A 																			&\text{Resolvent (2,8)}				\\
					&\text{10. } \neg A	 																&\text{Resolvent (1,5)}				\\
					&\text{11. } \Box																		&\text{Resolvent (9,10)}			\\
				\end{align*}		
								
				The claim is correct because $F$ is not fullfillable. In other words, a value for the variables in $F$ can be found to cause a contradiction. And because $F$ is the negation of the original claim, it is proven that the original claim must be correct.
				\end{proof}				
	\end{enumerate}
	
	
	
	
	\clearpage
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Fitch}
		\begin{enumerate}[a)] 
			\item									%Opgave 3a
				\textbf{Theorem}: $ \neg (A \wedge \neg C), C, A \wedge \neg B \vdash D $.
				
				\begin{figure}[!ht]
					\centering\begin{fitch}
					\fh \neg (A \wedge C )						& Hypothesis 1 \\
					\fa\fh C													& Hypothesis 2 \\
					\fa\fa\fh A \vee \neg B						& Hypothesis 3 \\
					\fa\fa\fa\fh \neg D								& Hypothesis 4 \\
					\fa\fa\fa\fa \neg (A \wedge C)		& rei 1				\\
					\fa\fa\fa\fa C										& rei 2				\\
					\fa\fa\fa\fa A \wedge \neg B			& rei 3				\\
					\fa\fa\fa\fa A										& $\wedge$ Elim 7 \\
					\fa\fa\fa\fa A \wedge C						& $\wedge$ intro 6,8 \\
					\fa\fa\fa\ \neg \neg D							& $\neg$ intro 4,5,9 \\
					\fa\fa\fa\ D											& $\neg$ elim 10
					
					\end{fitch}
					\caption{Fitch of  $ \neg (A \wedge \neg C), C, A \wedge \neg B \vdash D$.}
				\end{figure}
				
				
			\item 								%Opgave 3b
				\textbf{Theorem}: $ \neg ( \neg A \wedge \neg B ) \vdash A \vee B $.
				
				\begin{figure}[!ht]
					\centering\begin{fitch}
					\fh\neg(\neg A\wedge \neg B)				& Hypothesis 1 \\
					\fa\fh \neg(A \vee B)								& Hypothesis 2 \\
					
					\fa\fa\fh A  												& Hypothesis 3 \\
					\fa\fa\fa A \vee B									& $\vee$ intro. 3 \\
					\fa\fa\fa \neg(A \vee B)						& rei. 2 \\
					\fa\fa\ \neg A 											& $\neg$ intro. 3,4,5 \\
					
					\fa\fa\fh B  												& Hypothesis 4 \\
					\fa\fa\fa A \vee B									& $\vee$ intro. 7 \\
					\fa\fa\fa \neg(A \vee B)						& rei. 2 \\
					\fa\fa \neg B 										& $\neg$ intro. 7,8,9 \\
					
					\fa\fa\ \neg A \wedge \neg B				& $\wedge$ intro. 6,10 \\
					\fa\fa\ \neg(\neg A \wedge \neg B)	&  rei. 1 \\
					\fa\ \neg\neg(A \vee B)							& $\neg$ intro. 2,11,12 \\
					\fa\ A \vee B												& $\neg$ elim. 14 \\
					
					\end{fitch}
					\caption{Fitch of  $ \neg ( \neg A \wedge \neg B ) \vdash A \vee B $.}
				\end{figure}
				\clearpage
			\item 								%Opgave 3c
				\textbf{Theorem}: $ \vdash (A \rightarrow B) \vee (B \rightarrow A) $.
		
			\begin{figure}[!ht]
				\centering\begin{fitch}
				\fh \neg (( A \rightarrow B) \vee (B \rightarrow A))					& Hypothesis 1 \\
				
				\fa\fh A																											& Hypothesis 2 \\
				
				\fa\fa\fh B																										& Hypothesis 3 \\
				\fa\fa\fa A																										& rei 2 \\
				
				\fa\fa\ B \rightarrow A																				& $\rightarrow$ intro 3,4 \\
				\fa\fa\ (A \rightarrow B) \vee (B \rightarrow A)							& $\vee$ introcuction 5\\
				\fa\fa\ \neg ((A \rightarrow B) \vee (B \rightarrow A))				& rei. 1\\
				
				\fa\ \neg A																										& $\neg$ intro 2,6,7 \\
				
				\fa\fh A																											& Hypothesis 3 \\
				\fa\fa\fh \neg B																							& Hypothesis 4 \\
				\fa\fa\fa \neg A																							& rei 8\\
				\fa\fa\fa A																										& rei 9 \\
				\fa\fa\ \neg \neg B																						& $\neg$ intro 10,11,12 \\
				\fa\fa\ B																											& $\neg$ elim 13\\
				\fa\ (A \rightarrow B)																				& $\rightarrow$ intro 9,14 \\
				\fa\ (A \rightarrow B) \vee (B \rightarrow A)									& $\vee$ intro 15\\
				\fa\ \neg (A \rightarrow B) \vee (B \rightarrow A)						& rei 1\\
				
				\neg \neg (( A \rightarrow B) \vee (B \rightarrow A))					& $\neg$intro 1,16,17 \\
				(( A \rightarrow B) \vee (B \rightarrow A))									 	& $\neg$ elim 18	\\				
			
				\end{fitch}
				\caption{Fitch of  $ \vdash (A \rightarrow B) \vee (B \rightarrow A) $.}
			\end{figure}	
	
				
		\end{enumerate}
		
		
		
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Collections}
		\begin{enumerate}[a)] 
			\item									%Opgave 4a
				\textbf{Claim}. If $C \subseteq D^c $ then $ D \nsubseteq C^c $.
					The claim is not correct.\\
					\emph{Counterexample} \\
					Assume: $C = \left\{a,b\right\} $, $D^c = \left\{a,b,c\right\}$ and the univers $U=\left\{a,b,c,d,e\right\}$.	\\
					Then:
					$D=U-D^c = \left\{ d,e \right\} $ \\
					$C^c=U-C= \left\{ c,d,e \right\} $ \\
					This means that $ D \subseteq C^c $, but the claim is that $ D \nsubseteq C^c $.
					\clearpage
			\item 								%Opgave 4b
				\textbf{Claim}. If $C \nsubseteq D $ then $ D^c \nsubseteq C^c $.
				\begin{proof}\
				\\Assume: $ C \nsubseteq D$		\\
				To be proven: $ D^c \nsubseteq C^c $		
				
				Take a random $e \in C$. Due to the assumption  $ C \nsubseteq D$,	$e \notin D$ holds.			
	
				Then $e \in D^c$. And because $ e \in C$ it implies $e \notin C^c$.
				Because $x$ was choosen randomly it holds for all x.
				Therefore, the claim is correct.
				\end{proof}
			\item 							
	%Opgave 4c
				\textbf{Claim}. $ A \subseteq B $ desda $ A \cap B = A $.
				
				To be proven: $ (A \subseteq B) \Rightarrow (A \cap B = A) $ and $ (A \cap B = A) \Rightarrow (A \subseteq B) $
	
				\begin{proof}\
				\\ \emph{Claim a.} $(A \subseteq B) \Rightarrow (A \cap B = A) $\\
				Suppose that: $A \subseteq B$, which means that a random chosen $x$ which holds that $x\in A$ also holds that $x \in B$. \\
				To be proven: $(A \cap B) = A = \left\{ x | x \in A \wedge x \in B\right\}$ \\
				Suppose that: $(A \cap B)\neq A$ or $ A \neq \left\{x|x \in A \wedge x \in B\right\}$ \\
				Te be proven: $x \in A$ that is $x\notin B$	\\
				Take a random $x \in A$, according to $A \subseteq B$ $x$ must be $x \in B$. Which means that there is a contradiction so claim a must be true.
				
				\emph{Claim b.} $ (A \cap B = A) \Rightarrow (A \subseteq B) $\\
				Suppose that: $(A \cap B) = A = \left\{ x | x \in A \wedge x \in B\right\}$ \\
				To be proven: $A \subseteq B$, which means that a random chosen $x$ which holds that $x\in A$ also holds that $x \in B$. \\
				Suppose that: $A \nsubseteq B$ \\
				That means that there is a $x \in A$ that is not a $x \notin B$. But according to $(A \cap B) = A = \left\{ x | x \in A \wedge x \in B\right\}$
				$x \in A$ and $x \in B$. Which means that the $A \nsubseteq B$ is not true and claim b is true.
				\\\\
				Claim a and claim b are both true so the original claim is true as well.
				\end{proof}
				
			\item 								%Opgave 4d
				\textbf{Claim}. $ ( \mathscr{P}(A) \cup \mathscr{P}(B)) = \mathscr{P}(A \cup B) $.\\
				The claim is not correct.\\
				\emph{Counterexample} \\
				Assume: $A = \left\{ a,b \right\}$ and $B = \left\{ b,c \right\}$.\\
				Then:	\\
				\begin{align*}
					\mathscr{P}(A) 											&= \left\{ \left\{\right\}, \left\{a\right\} , \left\{b\right\} , \left\{a,b\right\} \right\}\\
					\mathscr{P}(B) 											&= \left\{ \left\{\right\}, \left\{b\right\} , \left\{c\right\} , \left\{b,c\right\} \right\} \\
					\mathscr{P}(A) \cup \mathscr{P}(B) 	&= \left\{ \left\{\right\}, \left\{a\right\}, \left\{b\right\}, \left\{c\right\}, \left\{a,b\right\}, \left\{b,c\right\} \right\} \\
					A \cup B 														&= \left\{a, b, c\right\} \\
					\mathscr{P}(A \cup B) 							&= \left\{ \left\{\right\}, \left\{a\right\}, \left\{b\right\}, \left\{c\right\}, \left\{a,b\right\}, \left\{b,c\right\}, \left\{a,c\right\}, \left\{a,b,c\right\} \right\} \\
				\end{align*}
				From the counterexample it is clearly visible that $ ( \mathscr{P}(A) \cup \mathscr{P}(B)) \neq \mathscr{P}(A \cup B) $
		\end{enumerate}

